Array Insertion at Middle Index in DSA Go - Time & Space Complexity

Time Complexity: Array Insertion at Middle Index

O(n)

Understanding Time Complexity

We want to understand how the time needed to insert an element in the middle of an array changes as the array grows.

Specifically, how does the work increase when the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "fmt"

func insertMiddle(arr []int, val int) []int {
    mid := len(arr) / 2
    arr = append(arr, 0)             // increase slice size by 1
    copy(arr[mid+1:], arr[mid:])     // shift elements right
    arr[mid] = val                   // insert value
    return arr
}

func main() {
    arr := []int{1, 2, 3, 4, 5}
    arr = insertMiddle(arr, 99)
    fmt.Println(arr)
}

This code inserts a value in the middle of an integer array by shifting elements to the right.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Shifting elements to the right using copy which moves elements from the middle to the end.
• How many times: The number of elements shifted is about half the array size (n/2) in the worst case.

How Execution Grows With Input

As the array size grows, the number of elements to shift grows roughly in proportion to the size.

Input Size (n)	Approx. Operations (shifts)
10	About 5
100	About 50
1000	About 500

Pattern observation: The work grows linearly with the size of the array.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert in the middle grows directly with the array size.

Common Mistake

[X] Wrong: "Inserting in the middle is always fast because we just put the value there."

[OK] Correct: We must move all elements after the middle to make space, which takes time proportional to how many elements are shifted.

Interview Connect

Understanding this helps you explain why arrays are not always the best choice for frequent middle insertions, showing your grasp of data structure trade-offs.

Self-Check

"What if we changed the array to a linked list? How would the time complexity of inserting in the middle change?"

Practice

1. What is the middle index of an array with 7 elements when inserting a new element?

easy

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Calculate middle index for odd length array

   For an array of length 7, middle index = length / 2 = 7 / 2 = 3 (integer division).

2. Step 2: Confirm insertion position

   The new element should be inserted at index 3 to be in the middle.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Middle index = 7/2 = 3 [OK]

Hint: Middle index = length divided by 2 (integer division) [OK]

Common Mistakes:

• Using length/2 + 1 for middle index
• Confusing zero-based index with count
• Using length/2 as float instead of integer

2. Which Go code snippet correctly inserts an element `x` at the middle index of slice `arr`?

easy

A. arr = append(arr, x)

B. arr = append(arr[:len(arr)/2], append([]int{x}, arr[len(arr)/2:]...)...)

C. arr = append(arr[:len(arr)/2], x, arr[len(arr)/2:])

D. arr = append(arr[:len(arr)/2+1], append([]int{x}, arr[len(arr)/2:]...)...)

Solution

1. Step 1: Understand slice insertion in Go

   To insert at middle, split slice at middle index, append new element slice, then append remaining slice.

2. Step 2: Check syntax correctness

   arr = append(arr[:len(arr)/2], append([]int{x}, arr[len(arr)/2:]...)...) correctly uses append with slices and variadic syntax to insert element at middle.

3. Final Answer:

   arr = append(arr[:len(arr)/2], append([]int{x}, arr[len(arr)/2:]...)...) -> Option B

4. Quick Check:

   Correct slice append syntax = arr = append(arr[:len(arr)/2], append([]int{x}, arr[len(arr)/2:]...)...) [OK]

Hint: Use append with slices and variadic ... to insert at middle [OK]

Common Mistakes:

• Trying to append element directly without wrapping in slice
• Incorrect index calculation with +1
• Appending element at end instead of middle

3. What is the output of this Go code?

arr := []int{1, 2, 3, 4}
mid := len(arr) / 2
arr = append(arr[:mid], append([]int{99}, arr[mid:]...)...)
fmt.Println(arr)

medium

A. [1 2 99 3 4]

B. [1 2 3 99 4]

C. [99 1 2 3 4]

D. [1 2 3 4 99]

Solution

1. Step 1: Calculate middle index

   len(arr) = 4, so mid = 4 / 2 = 2.

2. Step 2: Insert 99 at index 2

   arr[:2] = [1, 2], arr[2:] = [3, 4]. Insert 99 between them.

3. Step 3: Resulting array

   New arr = [1, 2, 99, 3, 4]

4. Final Answer:

   [1 2 99 3 4] -> Option A

5. Quick Check:

   Insert 99 at index 2 = [1 2 99 3 4] [OK]

Hint: Middle index = length/2; insert element at that index [OK]

Common Mistakes:

• Inserting element at wrong index
• Appending element at end instead of middle
• Misunderstanding slice append syntax

4. Identify the error in this Go code that tries to insert 50 at the middle of slice `nums`:

nums := []int{10, 20, 30, 40}
mid := len(nums) / 2
nums = append(nums[:mid], 50, nums[mid:])
fmt.Println(nums)

medium

A. Middle index calculation is wrong

B. Slice nums is not initialized

C. Cannot append element and slice together without wrapping element in slice

D. fmt.Println syntax error

Solution

1. Step 1: Check append syntax

   append(nums[:mid], 50, nums[mid:]...) tries to append int and slice variadically, which is invalid.

2. Step 2: Correct approach

   Wrap 50 in a slice: append(nums[:mid], append([]int{50}, nums[mid:]...)...).

3. Final Answer:

   Cannot append element and slice together without wrapping element in slice -> Option C

4. Quick Check:

   Append element + slice needs nested append [OK]

Hint: Wrap single element in slice when appending with another slice [OK]

Common Mistakes:

• Appending element and slice directly together
• Incorrect middle index calculation
• Forgetting variadic ... operator

5. Given a slice `data := []int{5, 10, 15, 20, 25}`, write the correct Go code to insert `100` exactly in the middle, then print the updated slice.

hard

A. mid := len(data)/2
data = append(data[:mid], 100, data[mid:])
fmt.Println(data)

B. mid := len(data)/2
data = append(data[:mid+1], append([]int{100}, data[mid:]...)...)
fmt.Println(data)

C. mid := len(data)/2
data = append(data, 100)
fmt.Println(data)

D. mid := len(data)/2
data = append(data[:mid], append([]int{100}, data[mid:]...)...)
fmt.Println(data)

Solution

1. Step 1: Calculate middle index

   len(data) = 5, so mid = 5 / 2 = 2 (integer division).

2. Step 2: Insert 100 at index 2 correctly

   Use nested append: append(data[:mid], append([]int{100}, data[mid:]...)...).

3. Step 3: Print updated slice

   Prints [5 10 100 15 20 25].

4. Final Answer:

   mid := len(data)/2 data = append(data[:mid], append([]int{100}, data[mid:]...)...) fmt.Println(data) -> Option D

5. Quick Check:

   Insert at middle with nested append = mid := len(data)/2 data = append(data[:mid], append([]int{100}, data[mid:]...)...) fmt.Println(data) [OK]

Hint: Use nested append with slices to insert element at middle [OK]

Common Mistakes:

• Using mid+1 instead of mid for insertion
• Appending element at end instead of middle
• Appending element and slice without wrapping element